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BBMP1103 – Chapter 2 - Richard Ng (2008) Page 1 of 47
BBMP 1103BBMP 1103
Matematik PengurusanMatematik Pengurusan
Chapter:2 Chapter:2 –– MatrixMatrix
Prepared by
Richard NgRichard Ng
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 2 of 47
1. Introduction to Matrix
Matrix consists of data arranged in the form of rows andcolumns
Examples:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
251
⎥⎦
⎤⎢⎣
⎡5142
⎥⎦
⎤⎢⎣
⎡10
21
53
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
327510321
[ ]31⎥⎦
⎤⎢⎣
⎡23
1x2 Matrix 2x1 Matrix 2x2 Matrix
3x1 Matrix 2x3 Matrix 3x3 Matrix
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 3 of 47
2. Elements of a 3 x 3 Matrix
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
333231
232221
131211
aaaaaaaaa
a11 is located at row one and column onea12 is located at row one and column twoa13 is located at row one and column threea21 is located at row two and column onea22 is located at row two and column twoa32 is located at row three and column twoaij is located at row i and column j
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 4 of 47
3. Classifications of Matrices
⎥⎦
⎤⎢⎣
⎡23
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
102
⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢
⎣
⎡
−0112
Column Matrix2x1
Column Matrix3x1
Column Matrix4x1
[ ]21 [ ]102 − [ ]4013Row Matrix
1x2Row Matrix
1x3Row Matrix
1x4
Column Matrix (Column Vector): has only 1 column
Row Matrix (Row Vector): has only 1 row
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 5 of 47
⎥⎦
⎤⎢⎣
⎡0000
⎥⎦
⎤⎢⎣
⎡000000
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
000000000
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
000
000
⎥⎦
⎤⎢⎣
⎡4321
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
021843512
2x2 Matrix 3x3 Matrix
Zero Matrix (Null Matrix): all elements are zero
Square Matrix: number rows = number of columns
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 6 of 47
Main Diagonal
Elements located on the main diagonal in a squarematrix are elements which have the same row and column number. Examples:
In example (i): a11 = 1 and a22 = 4 hence the elements on the main diagonal are 1 and 4
In example (ii): a11 = 2, a22 = 4 dan a33 = 0 hence the elements on the main diagonal are 2, 4 dan 0
⎥⎦
⎤⎢⎣
⎡4321
(i) ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
021843512
(ii)⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
333231
232221
131211
aaaaaaaaa
(iii)
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 7 of 47
An Identity matrix is a square matrix where all the elements on the main diagonal = 1 and the rest = 0
⎥⎦
⎤⎢⎣
⎡1001
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
100010001
Identity Matrix
Example:
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 8 of 47
4. Matrix OperationsTranspose
Elements in a row elements in a columnElements in a column elements in a row
Example: 1
If ⎥⎦
⎤⎢⎣
⎡=
142301
A Then
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
142
301
AT
Example: 2
If⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
987654321
B Then⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
963852741
BT
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 9 of 47
Matrix Addition
Addition in a matrix involves elements at the same position and same type of matrix only
Example: 3
⎥⎦
⎤⎢⎣
⎡+⎥
⎦
⎤⎢⎣
⎡3102
4321
⎥⎦
⎤⎢⎣
⎡=
7423
Example: 4
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
231
412
013
201
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
244
613
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 10 of 47
Example: 5
Example: 6
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡−
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
231
412
013
201
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−−−
=222
211
⎥⎦
⎤⎢⎣
⎡−⎥
⎦
⎤⎢⎣
⎡3102
4321
⎥⎦
⎤⎢⎣
⎡−=
1221
Matrix Subtraction
Subtraction in a matrix involves elements at the same position and same type of matrix only
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 11 of 47
Scalar multiplication – when a matrix is multiplied by a scalar, every element must be multiplied with that scalar
Example: 7
⎥⎦
⎤⎢⎣
⎡=
4321
AIf Then ⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡=
8642
4321
22A
Example: 8
If⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
013412201
B Then⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
0391236603
013412201
33B
Matrix Multiplication
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 12 of 47
Multiplication of Matrices
Tips:
Matrix2 x 1
x Matrix1 x 2
Matrix2 x 2
Matrix2 x 3
x Matrix3 x 3
Matrix2 x 3
Multiplication of Matrices is only possible when the number of rows in the 1st matrix = number of columns in the 2nd
Matrix
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 13 of 47
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡++++
=⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡10352
46031402
1021
4312
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡++++
=⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡9504
90320004
3102
3102
3102 2
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡++
=⎥⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡85
4432
12
4231
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
++++++++++++++++++
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
47532511412
004106104030200230092103192
011303232
102210131
Example: 9
Example: 10
Example: 11
Example: 12
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 14 of 47
a) For a 2 x 2 Matrix
⎥⎦
⎤⎢⎣
⎡3412
5. Determinant
Is defined only for square matricesIs used to find the invertible matrix
Finding determinant using cross multiplication
If A =
Hence the determinant of A = |A| = (2)(3) – (4)(1)
= 6 – 4= 2
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 15 of 47
b) For a 3 x 3 Matrix
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
024130312
402
231
If B =
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
024130312
Hence |B| =
= [(2)(3)(0)+(1)(1)(4)+(3)(0)(2)]–[(4)(3)(3)+(2)(1)(2)+(0)(0)(1)]
= [0 + 4 + 0]–[36 + 4 + 0]
= -36
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 16 of 47
If A = ⎥⎦
⎤⎢⎣
⎡ −4210
and B = ⎥⎦
⎤⎢⎣
⎡1021
Find:
a) |A| b) |B| c) |AB| d) |BA|
e) |A||B| f) |B||A|
Example: 13
Answer:
Hence |A|⎥⎦
⎤⎢⎣
⎡ −4210
a) A = = (0)(4) – (2)(-1)
= 0 – (-2)
= 2
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 17 of 47
b) B = ⎥⎦
⎤⎢⎣
⎡1021
c) AB = ⎥⎦
⎤⎢⎣
⎡ −4210
⎥⎦
⎤⎢⎣
⎡1021
= ⎥⎦
⎤⎢⎣
⎡++−++4402
)1(000
Hence |AB|
= ⎥⎦
⎤⎢⎣
⎡ −8210
Hence |B|= (1)(1) – (0)(2)
= 1 – 0
= 1
= ⎥⎦
⎤⎢⎣
⎡ −8210
= (0)(8) – (2)(-1)
= 0 – (-2) = 2
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 18 of 47
d) BA = ⎥⎦
⎤⎢⎣
⎡1021
⎥⎦
⎤⎢⎣
⎡ −4210
= ⎥⎦
⎤⎢⎣
⎡+++−+40208140
= ⎥⎦
⎤⎢⎣
⎡4274
= 2
= (4)(4) – (2)(7)
Hence |BA|
= 16 – 14
⎥⎦
⎤⎢⎣
⎡4274
=
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 19 of 47
e) From (a) dan (b):
|A| = 2 |B| = 1
Hence |A||B| = (2)(1) = 2
f) From (a) dan (b):
|A| = 2 |B| = 1
Hence |B||A| = (1)(2) = 2
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 20 of 47
Given P = ⎥⎦
⎤⎢⎣
⎡4321
Step: 1 – Find the matrix minor of P:
Minor of P:
⎥⎦
⎤⎢⎣
⎡=
4321
11a 4= ⎥⎦
⎤⎢⎣
⎡=
4321
12a 3=
⎥⎦
⎤⎢⎣
⎡=
4321
21a 2= ⎥⎦
⎤⎢⎣
⎡=
4321
22a 1=
⎥⎦
⎤⎢⎣
⎡=
2221
1211
aaaa
⎥⎦
⎤⎢⎣
⎡=
1234
Example: 14
Finding determinant using Cofactor Method
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 21 of 47
Step: 2 – Find the cofactor of P:
Element in a cofactor = element in minor x (-1)(i+j)
Hence cofactor of P ⎥⎦
⎤⎢⎣
⎡
−×−×−×−×
=++
++
)22()12(
)21()11(
)1(1)1(2)1(3)1(4
⎥⎦
⎤⎢⎣
⎡
−×−×−×−×
=)4()3(
)3()2(
)1(1)1(2)1(3)1(4
⎥⎦
⎤⎢⎣
⎡−
−=
1234
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 22 of 47
Step: 3 – Find the determinant:
Determinant = Product of elements in the original matrix and elements of the cofactor of a row
Original matrix P = ⎥⎦
⎤⎢⎣
⎡4321
Cofactor of P ⎥⎦
⎤⎢⎣
⎡−
−=
1234
Choose first row to find the determinant:
Determinant = (1)(4) + (2)(-3)
= 4 + (-6)
= -2⎥⎦
⎤⎢⎣
⎡4321
⎥⎦
⎤⎢⎣
⎡−
−1234
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 23 of 47
Example: 15
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
024130312
Step: 1 – Find the elements in the 2nd row of Minor Q:
Given that Q =
Using the second row to find the determinant, then:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
024130312
21a = (1)(0) – (2)(3) = 0 – 6 = - 6
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 24 of 47
= (2)(0) – (4)(3) = 0 – 12 = - 12⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
024130312
22a
= (2)(2) – (4)(1) = 4 – 4 = 0⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
024130312
23a
Step: 2 – Find elements in the 2nd row of cofactor Q:
6)1)(6()1)(6( )12(21 =−−=−−= +a
12)1)(12()1)(12( )22(22 −=−=−−= +a
0)1)(0()1)(0( )32(23 =−=−= +a
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 25 of 47
Step: 3 – Find the determinant:
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
024130312
6 -12 0
Determinant = (0)(6) + (3)(-12) + (1)(0)
= 0 + (-36) + (0)
= -36
Determinant = Product of elements in the original matrix and elements of the cofactor of a row
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 26 of 47
6. Inverse Matrix
⎥⎦
⎤⎢⎣
⎡3412
Let A =
Step: 1 – Find the determinant
= (2)(3) – (4)(1)
= 6 – 4
= 2
Finding the Inverse Matrix of the type 2x2
⎥⎦
⎤⎢⎣
⎡3412
|A| =
Example: 16
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 27 of 47
Step: 3 – Find the inverse matrix of A = A-1
A-1 = ⎥⎦
⎤⎢⎣
⎡−
−2413
x|A|
1
= ⎥⎦
⎤⎢⎣
⎡−
−2413
x21
⎥⎥
⎦
⎤
⎢⎢
⎣
⎡
−
−
1221
23
=
Step: 2 – Find the adjoint of A
⎥⎦
⎤⎢⎣
⎡3412
⎥⎦
⎤⎢⎣
⎡−
−2413 Elements of Main Diagonal
⇒Change positionElements Second Diagonal⇒Change sign
[ ]AintAdjox|A|
1A 1 =−
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 28 of 47
Example: 17
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
Let A =
Step: 1 – Find the determinant of A
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
|A| =324
201
−= [6+0+0] – [4+0+0]
= 6 – 4= 2
Finding the Inverse Matrix of the type 3x3
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 29 of 47
Step: 2 – Find the Minor for matrix A
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a11 = = (2)(3) – (-3)(0) = 6 – 0 = 6
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a12 = = (0)(3) – (2)(0) = 0 – 0 = 0
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a13 = = (0)(-3) – (2)(2) = 0 – 4 = -4
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 30 of 47
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a21 = = (4)(3) – (-3)(1) = 12 + 3 = 15
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a22 = = (1)(3) – (2)(1) = 3 – 2 = 1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a23 = = (1)(-3) – (2)(4) = -3 – 8 = -11
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 31 of 47
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a31 = = (4)(0) – (2)(1) = 0 - 2 = -2
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a32 = = (1)(0) – (0)(1) = 0 – 0 = 0
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
a33 = = (1)(2) – (0)(4) = 2 – 0 = 2
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 32 of 47
Matrix Minor A =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
20211115406
Matrix cofactor A =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
20211115406
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
+−+−+−+−+
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
20211115406
=
Step: 3 – Find the cofactor of A
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 33 of 47
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−−
21140102156
Step: 4 – Find the adjoint of A
Adjoint A = [Cofactor A]T
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
20211115406
A-1 = = =
Step: 5 – Find A-1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−−
21140102156
|A|1
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−−
21140102156
21
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
12112
0210
1215
3
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 34 of 47
7. Solving Simultaneous Equations
Example:18Find the value of x and y for the following equations:
x + y = 22x –y = 1
Step: 1 – Change the equations in the form AX = Bx + y = 22x –y = 1 ⎥
⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡− 1
21211
yx
Let A = ⎥⎦
⎤⎢⎣
⎡−1211
Hence |A| = (-1) – (2) = -3
Step: 2 – Find the determinant of A
Inverse Matrix Method
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 35 of 47
Step: 3 – Find the adjoint matrix A
⎥⎦
⎤⎢⎣
⎡−1211
⎥⎦
⎤⎢⎣
⎡−
−−1211
Step: 4 – Find the inverse matrix of A
A-1 = ⎥⎦
⎤⎢⎣
⎡−
−−1211
||1A = ⎥
⎦
⎤⎢⎣
⎡−
−−12111
3- =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−31
32
31
31
Step: 5 – Multiply B with the inverse matrix
⎥⎦
⎤⎢⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−=⎥
⎦
⎤⎢⎣
⎡12
31
32
31
31
yx
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 36 of 47
Hence ⎥⎦
⎤⎢⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
+=⎥
⎦
⎤⎢⎣
⎡11
3333
31
34
31
32
yx
Thus x =1 and y = 1
x + y = 2 ……(a)2x –y = 1 ……(b)
(a) + (b) => 3x = 3Then x = 1
Substitute x = 1 into (a) Then 1 + y = 2and y = 1
Check:
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 37 of 47
x + 4y + z = 12y = 3
2x – 3y + 3z = 0
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
− 332020141
Step: 1 – Change the equations in the form AX = B
x + 4y + z = 12y = 3
2x – 3y + 3z = 0 ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
zyx
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
031
Step: 2 – Find the determinant of A
Dari ms 28: |A| = 2
Example: 19
Solve the following equations:
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 38 of 47
Step: 3 – Find the minor of A
From slide 32: Minor A = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−−
20211115406
Step: 4 – Find the cofactor of A
From slide 32: Cofactor A = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
20211115406
Step: 5 – Find the adjoint of A
From slide 33: Adjoint A =⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−
−−
21140102156
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 39 of 47
Step: 6 – Find the inverse matrix A-1
From slide 33: Inverse A =
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
12112
0210
12
153
Step: 7 – Solve the equations
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
031
AMatrixInverse
zyx
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 40 of 47
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡−
=
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
++−
++
+⎟⎠⎞
⎜⎝⎛−+
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥⎥⎥⎥
⎦
⎤
⎢⎢⎢⎢⎢⎢
⎣
⎡
−
−−
=⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
2114211
2119
02332
0230
02453
031
12112
0210
12
153
zyx
Hence the value of x = 2119−
y = 211
and z =2114
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 41 of 47
Solve the equations:
5202
=−=+
yxyx
Step: 1 – Change the equation in the form AX = B
⎥⎦
⎤⎢⎣
⎡=⎥
⎦
⎤⎢⎣
⎡⎥⎦
⎤⎢⎣
⎡− 5
01221
yx
Step: 2 – Find |A|, |A1| dan |A2|
Let A = ⎥⎦
⎤⎢⎣
⎡−1221
|A| = (-1) – (4) = -5
Example: 20
Cramer’s Rule Method
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 42 of 47
Let A1 = ⎥⎦
⎤⎢⎣
⎡−1520
|A1| = (0) – (10) = -10
Let A2 = ⎥⎦
⎤⎢⎣
⎡5201
|A2| = (5) – (0) = 5
Step: 3 – Find the value of x and y
x =|A||A| 1 =
5-10- = 2
y =|A||A| 2 = = -1
5-5
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 43 of 47
Solve the equations:
02242332
=−=−+=+−
zxzyxzyx
Step: 1 – Change the equations in the form AX = B
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡=
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
043
202121312
zyx
Example: 21
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 44 of 47
Step: 2 – Find |A|, |A1|, |A2| dan |A3|
⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
202121312
Let A =
Hence |A| = [-8+2+0] – [12+0+2] = -6 –14 = -20
212
021−
Let A1 = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−
−
200124313
043
021−
Hence |A1| = [-12+0+0] – [0+0+8] = -12 – 8 = -20
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 45 of 47
Let A2 = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡
−−202141332
212
043
Hence |A2| = [-16+(-6)+0] – [24+0+(-6)] = -22 – 18 = -40
Let A3 = ⎥⎥⎥
⎦
⎤
⎢⎢⎢
⎣
⎡ −
002421312
212
021−
Hence |A3| = [0+(-8)+0] – [12+0+0] = -8 – 12 = -20
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 46 of 47
Step: 3 – Find the value of x, y and z
x =|A||A| 1 = = 1
2020
−−
y = = = 2|A||A| 2
2040
−−
z = = = 1|A||A| 3
2020
−−
BBMP1103 – Chapter 2 - Richard Ng (2008) Page 47 of 47
End ofChapter
2